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Question

Find the area of the smaller region bounded by the ellipse x29+y24=1 and the line x3+y2=1.

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Solution

Given the curve (represents an elliplse with centre at (0, 0)

x29+y24=1 ...(i)

and equation of line x3+y2=1 ...(ii)

For the points of intersection of ellipse and line put the value of x from

Eq. (ii) in Eq. (i). we get (y21)2+y24=1

y24+1y+y24=1y22y=0y=0,2

When y = 0, x = 3 and y = 2, x= 0

i.e., intersection points are A(3, 0) and B(0, 2)

Required area = (Area under the curve x29+y24=1, between x = 0 and x = 3) - (Area under the line x3+y2=1, between x = 0 and x = 3)

=3021x29dx302(1x3)dx=233032x2dx2330(3x)dx=23[x232x2+92sin1x3]3023[3xx22]30
[a2x2dx=x2a2x2+a22sin1xa]

=23[0+92sin1(1)0]23[9920]=3(π2)(3)=3(π21)=32(π2)sq unit.

=23[0+92sin1(1)0]23[9920]=3(π2)(3)=3(π21)=32(π2)sq unit.


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